Logarithmic Functions MCQ Quiz - Objective Question with Answer for Logarithmic Functions - Download Free PDF

Concept:

• log A + log B = log AB
• log A - log B = log(A/B)
• log a^k = k log a
• $\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$
• $\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$

Calculation:

Given,

${\displaystyle \frac{logx+log{x}^4+log{x}^9+....+log{x}^{n^2}}{logx+log{x}^2+log{x}^3+....+log{x}^n}}$

⇒ ${\displaystyle \frac{log[x(x^4)(x^9)....(x^{n^2})]}{log[x(x^2)(x^3)...(x^n)]}}$

⇒ ${\displaystyle \frac{log[x^{1+4+9+...n^2}]}{log[x^{1+2+3+...+n}]}}$

⇒ ${\displaystyle \frac{log[x^{\sum _{k=1}^n{k}^2}]}{log[x^{\sum _{k=1}^{n}k}]}}$

⇒ ${\displaystyle \frac{\sum _{k=1}^n{k}^{2}log[x]}{\sum _{k=1}^{n}klog[x]}}$

⇒ ${\displaystyle \frac{\sum _{k=1}^n{k}^2}{\sum _{k=1}^{n}k}}$

⇒ ${\displaystyle \frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}}$

⇒ $\frac{2n+1}{3}$

∴ The correct answer is option (1).

Given:

8k = 2

Formula used:

log_b a = c ^{if and only if} b^c = a

Calculation:

Given: 8k = 2

⇒ log₈ 2 = x

∴ The correct answer is option (3).

Given:

log₁₀ 10000 = ?

Formula used:

log_b a = c ⟺ b^c = a

Calculation:

log₁₀ 10000 = ?

10000 = 10⁴

⇒ log₁₀ 10000 = 4

∴ The correct answer is option (4).

Explanation -

We have,

$4^x+2^{2x-1}=3^{x+\frac{1}{2}}+3^{x-\frac{1}{2}}$

⇒ 2 × 2^2x-1 + 2^2x-1 = $3^{x-\frac{1}{2}}$ × 3 + $3^{x-\frac{1}{2}}$

⇒ 2^2x-1 (2 + 1) = $3^{x-\frac{1}{2}}$ (3 + 1)

⇒ 22x-1 × 3 = $3^{x-\frac{1}{2}}$ × 4

⇒ 22x-3 = $3^{x-\frac{3}{2}}$

⇒ ${\left(2^2\right)}^{x-\frac{3}{2}}$ = $3^{x-\frac{3}{2}}$ 

⇒ $4^{x-\frac{3}{2}}$ = $3^{x-\frac{3}{2}}$ 

⇒ $x-\frac{3}{2}$ = 0 

⇒ $x=\frac{3}{2}$

​Hence Option (2) is correct.

Concept:

Logarithm properties:  

Product rule: The log of a product equals the sum of two logs.

${\log }_{\mathrm{a}}\left(\mathrm{m}\mathrm{n}\right)={\log }_{\mathrm{a}}\mathrm{m}+{\log }_{\mathrm{a}}\mathrm{n}$

Quotient rule: The log of a quotient equals the difference of two logs.

${\log }_{\mathrm{a}}\frac{\mathrm{m}}{\mathrm{n}}={\log }_{\mathrm{a}}\mathrm{m}-{\log }_{\mathrm{a}}\mathrm{n}$

Power rule: In the log of power the exponent becomes a coefficient.

${\log }_{\mathrm{a}}{\mathrm{m}}^{\mathrm{n}}=\mathrm{n}{\log }_{\mathrm{a}}\mathrm{m}$

Formula of Logarithms:

If $\mathrm{l}\mathrm{o}{\mathrm{g}}_{\mathrm{a}}\mathrm{x}=\mathrm{b}$ then x = ab (Here a ≠ 1 and a > 0)

Calculation:

Given: ${\log }_3({\mathrm{x}}^4-{\mathrm{x}}^3)-{\log }_3(\mathrm{x}-1)=3$

$\Rightarrow {\log }_3\left[\frac{({\mathrm{x}}^4-{\mathrm{x}}^3)}{(\mathrm{x}-1)}\right]=3$        (∵ ${\log }_{\mathrm{a}}\frac{\mathrm{m}}{\mathrm{n}}={\log }_{\mathrm{a}}\mathrm{m}-{\log }_{\mathrm{a}}\mathrm{n}$)

$\Rightarrow {\log }_3\left[\frac{{\mathrm{x}}^3(\mathrm{x}-1)}{(\mathrm{x}-1)}\right]=3$

$\Rightarrow {\log }_3{\mathrm{x}}^3=3$

$\Rightarrow 3{\log }_3\mathrm{x}=3$               (∵ ${\log }_{\mathrm{a}}{\mathrm{m}}^{\mathrm{n}}=\mathrm{n}{\log }_{\mathrm{a}}\mathrm{m}$) 

$$\begin{gathered} \Rightarrow {\log }_3\mathrm{x}=1 \\ \therefore \mathrm{x}=3 \end{gathered}$$

Concept:

$a^b=x\iff lo{g}_{a}x=b$, where $a\ne 1$ and a > 0 and x be any number.

Calculation:

Given: 92^1/5 = 4.

As we know that, $a^b=x\iff lo{g}_{a}x=b.$

Comparing 92^1/5 = 4 with $a^b=x$ we have,

Here, a = 92, b = 1 / 5 and x = 4.

Concept:

Logarithm properties

1. Product rule: The log of a product equals the sum of two logs.

${\log }_a\left(mn\right)={\log }_{a}m+{\log }_{a}n$

2. Quotient rule: The log of a quotient equals the difference of two logs.

${\log }_a\frac{m}{n}={\log }_{a}m-{\log }_{a}n$

3. Power rule: In the log of a power the exponent becomes a coefficient.

${\log }_a{m}^n=n{\log }_{a}m$

4. Change of base rule

${\log }_{m}n=\frac{{\log }_{a}n}{{\log }_{a}m}$

If m = n;
⇒ ${\log }_{m}m=\frac{{\log }_{a}m}{{\log }_{a}m}=1$

5. ${\log }_{m}n=\frac{1}{{\log }_{n}m}$

Calculation:

Here, we have to find the value of ${\log }_7{\log }_7\sqrt{7\sqrt{7\sqrt{7}}}$

${\log }_7{\log }_7\sqrt{7\sqrt{7\sqrt{7}}}$

= log₇ log₇ (7^1/2 × 7^1/4 × 7^1/8)

= log₇ log₇ (7(^{1/2 + 1/4 + 1/8)})

= log₇ log₇ (7^{(4 + 2 + 1)/8})

= log₇ log₇ (7^7/8)

From power rule;

= log₇ (7/8) log₇7

= log₇ (7/8) × 1 = log₇ (7/8) = log₇ 7 – log₇ 8

= 1 – log₇ 8 = 1 – log₇ 2³

Concept:

Logarithm properties:  

Product rule: The log of a product equals the sum of two logs.

${\log }_{\mathrm{a}}\left(\mathrm{m}\mathrm{n}\right)={\log }_{\mathrm{a}}\mathrm{m}+{\log }_{\mathrm{a}}\mathrm{n}$

Quotient rule: The log of a quotient equals the difference of two logs.

${\log }_{\mathrm{a}}\frac{\mathrm{m}}{\mathrm{n}}={\log }_{\mathrm{a}}\mathrm{m}-{\log }_{\mathrm{a}}\mathrm{n}$

Power rule: In the log of power the exponent becomes a coefficient.

${\log }_{\mathrm{a}}{\mathrm{m}}^{\mathrm{n}}=\mathrm{n}{\log }_{\mathrm{a}}\mathrm{m}$

Formula of Logarithms:

If $\mathrm{l}\mathrm{o}{\mathrm{g}}_{\mathrm{a}}\mathrm{x}=\mathrm{b}$ then x = ab (Here a ≠ 1 and a > 0)

Calculation:

Given: ${\log }_4({\mathrm{x}}^2-1)-{\log }_4(\mathrm{x}+1)=1$

$\Rightarrow {\log }_4\left[\frac{({\mathrm{x}}^2-1)}{(\mathrm{x}+1)}\right]=1$        (∵ ${\log }_{\mathrm{a}}\frac{\mathrm{m}}{\mathrm{n}}={\log }_{\mathrm{a}}\mathrm{m}-{\log }_{\mathrm{a}}\mathrm{n}$)

$\Rightarrow {\log }_4\left[\frac{(\mathrm{x}-1)(\mathrm{x}+1)}{(\mathrm{x}+1)}\right]=1$

$\Rightarrow {\log }_4(\mathrm{x}-1)=1$

⇒ (x - 1) = 4

∴ x = 5

Concept:

Logarithms:

• If ab = x, then we say that loga x = b.
• log_a a = 1.
• log_a (xy) = log_a x + log_a y.

Calculation:

We know that 80 = 23 × 10.

Changing the given logarithms to log 2, we get:

log₁₀ 80

= log₁₀ (23 × 10)

= log10 2³ + log10 10

= 3 (log10 2) + 1

= 3(0.3010) + 1

= 1.9030.

Given:

5x-1 = (2.5)log105

Formula Used:

If a^x = n then x = log_aⁿ

log_a^b = log_e^b/log_e^a

Calculation:

We have 5x-1 = (2.5)log105

⇒ (2.5)log105 = 5x-1 

⇒ log₁₀⁵  = log_2.5^5x-1 

⇒ log105  = (x - 1) log2.55

⇒ (x - 1) = (log105)/(log2.55)

⇒ (x - 1) = log₁₀^2.5

⇒ x = log102.5 _{+ 1}

⇒ x = log102.5 ₊ log₁₀¹⁰

⇒ x = log1010 × 2.5

⇒ x = log1025

⇒ x = log105²

⇒ x = 2log10​5

∴ The value of x is 2log10​5.

Concept:

Logarithm properties

1. Product rule: The log of a product equals the sum of two logs.

${\log }_a\left(mn\right)={\log }_{a}m+{\log }_{a}n$

2. Quotient rule: The log of a quotient equals the difference of two logs.

${\log }_a\frac{m}{n}={\log }_{a}m-{\log }_{a}n$

3. Power rule: In the log of a power the exponent becomes a coefficient.

${\log }_a{m}^n=n{\log }_{a}m$

4. Change of base rule

${\log }_{m}n=\frac{{\log }_{a}n}{{\log }_{a}m}$

If m = n;

⇒ ${\log }_{m}m=\frac{{\log }_{a}m}{{\log }_{a}m}=1$

5. ${\log }_{m}n=\frac{1}{{\log }_{n}m}$

Calculation:

Here, we have to find the value of ${\log }_3{\log }_3\sqrt{3\sqrt{3}}$

Now,

${\log }_3{\log }_3\sqrt{3\sqrt{3}}$ = log₃ log₃ (3^1/2 × 3^1/4)

= log₃ log₃ (3^{(1/2 + 1/4)})

= log₃ log₃ (3^{(2 + 1)/4})

= log₃ log₃ (3^3/4)

From power rule;

= log₃ [(3/4)× log₃3]                [∵ log_a (m) ⁿ = n × log_a (m)]

= log₃ (3/4)                  (∵ log_m m = 1)

= log₃ (3/4) = log₃ 3 – log₃ 4

= 1 – log₃ 4 = 1 – log₃ 2²

= 1 – 2 log₃ 2

Concept:

Formula used:

• ${\log }_{a}b=\frac{1}{{\log }_{b}a}$
• Log_a M + log_a N = log_a (MN)

Factorial:

• n! = 1 × 2 × 3 × ⋯ × (n – 1) × n

Calculation:

Using ${\log }_{a}b=\frac{1}{{\log }_{b}a}$

$\frac{1}{{\log }_2\mathrm{N}}+\frac{1}{{\log }_3\mathrm{N}}+\dots +\frac{1}{{\log }_{100}\mathrm{N}}={\log }_{\mathrm{N}}2+{\log }_{\mathrm{N}}3+\dots +{\log }_{\mathrm{N}}100$

= log_N (2 × 3 × ⋯ × 100)

= log_N (100!)

$=\frac{1}{{\log }_{100!}N}$

Concept:

Logarithm Rule

log mⁿ = n log m

Calculations:

Let x, y, z are three consecutive positive integers.

⇒ y = x + 1 and z = y + 1

⇒ z = x + 2

Consider, log (1 + xz)

= log [1 + x(x+2)]

= log [1 + x² + 2x]

= log (1 + x)²

= 2 log (1 + x)

= 2 log y

Hence, If x, y, z are three consecutive positive integers, then log (1 + xz) is 2 log y

CONCEPT:

• By base changing theorem we know that ${\log }_{b}a=\frac{{\log }_{x}a}{{\log }_{x}b}=\frac{1}{{\log }_{a}b}$.
• log _x (x) = 1

CALCULATION:

Given that $\left\{\frac{1}{{\log }_{9}60}+\frac{1}{{\log }_{16}60}+\frac{1}{{\log }_{25}60}\right\}$

By base changing we can write it as - 

⇒ log₆₀9 + log6016 +log6025

By product rule of logarithms we can write it again as - 

⇒ log₆₀(9 x 16 x 25) = log₆₀(3600)

⇒ log₆₀ (60)² = 2 log₆₀(60) = 2

Therefore, option (3) is the correct answer.